Here's the problem that insprired my question:

Suppose $X$ is the set of real numbers, $\mathcal B$ is the Borel $\sigma$-algebra, and $m$ and $n$ are two measures on $(X, \mathcal B)$ such that $m((a,b)) = n((a,b)) < \infty$ whenever $-\infty < a< b < \infty$. Prove that $m(A) = n(A)$ whenever $A \in \mathcal B$.

When I look at this, I want to say that this problem is essentially trivial, but I can't convince myself that it actually is. I want to say that since these measures agree on sets that generate $\mathcal B$, and every $\mathcal B$-measurable set can be formed by taking countable intersections and unions of these types of sets, then the result follows from the fact that these are both measures.

So my questions are:

  1. Can we actually say that every Borel measurable set can be written as a countable union or intersection of these finite open intervals?

  2. Does the result immediately follow from properties of measures?

  • 2
    $\begingroup$ I think that this might help you. A consequence of the monotone class theorem is that one can prove that two measures coincide on a set stable under finite intersection and that generates your borel sigma algebra. This corollary (unfortunately is in french) helps you understand better what i mean. $\endgroup$
    – Bman72
    May 28, 2014 at 16:11
  • $\begingroup$ From what I can tell, the proof of the first part is essentially what I'm after. It looks like I need to first show $\mathcal G$ is a monotone class and lastly show that $\mathcal G = \mathcal B$ so that $\mu = \nu$ on all of the Borel measurable sets. My french isn't great, but I think I can decipher. Thanks! $\endgroup$
    – dannum
    May 28, 2014 at 16:55
  • $\begingroup$ If i'm not wrong (i should check but I'm quite sure) from the definition of monotone class you can see that every sigma algebra is a monotone class hence $\sigma(C) \subset \mathcal{M}(C)$. If you can find a set which is stable under finite intersection that generates your sigma algebra then you have that $\mathcal{M}(C) = \sigma(C)$ hence since $\mathcal{M}(C) \subset \mathcal{G} \subset \sigma(C)$ you have that your set G is exactly the sigma algebra and hence every set in the sigma algebra can be measured by a unique measure :) $\endgroup$
    – Bman72
    May 28, 2014 at 18:06
  • $\begingroup$ What confuses me still is that they are applying the monotone class theorem here, but our set $\mathcal C = \{(a,b): a,b \in \mathbb R\}$ is not an algebra as far as I can tell. It's closed under intersection, but I don't think it's closed under complementation. I assume the finiteness of $m((a,b)) = n((a,b))$ takes places of some condition that we need to apply the monotone class theorem? $\endgroup$
    – dannum
    May 28, 2014 at 19:47
  • $\begingroup$ If you look at what the corollary tells you the only thing that you need, to confirm that your two measures coincide, is that your set $\mathcal{C} := \{(a,b): a,b \in \mathbb{R} \}$ must be closed under finite intersection, i.e if $A, B \in \mathcal{C} \implies A \cap B \in \mathcal{C}$ (that is actually true), and that it generates your sigma algebra, in our case $\mathcal{B}(\mathbb{R})$ (this is also the case). Now you have that since your set have finite measure (or the all space can be generated by finite measurable sets) your measure coincide, just by applying the corollary. $\endgroup$
    – Bman72
    May 28, 2014 at 20:22

4 Answers 4


You can also look in to the Dynkin's $\pi - \lambda$ Theorem, which is equivalent with monotone class theorem.

Definition Let $P$ and $L$ be collections of subsets of a set $X$,

$P$ is a $\pi$-system if it is closed under finite intersections.

$L$ is a $\lambda$-system if the following hold:

  1. $\emptyset \in L$;
  2. if $A\in L$ then $A^c \in L$;
  3. $L$ is closed under countable disjoint unions.

Dynkin's $\pi - \lambda$ Theorem: Let $P$ be a $\pi$-system of subsets of $X$ and $L$ a $\lambda$-system of subsets of $X$. Suppose that $P\subset L$, then: $$\sigma(P) \subset L.$$

How to apply it:

  1. $P:=\{(a,b) : -\infty < a<b<\infty\}$ forms a $\pi$-system, and in general the collection of intervals (or boxes in higher dimensions) always forms a $\pi$-system.

  2. $L:=\{A\in \mathcal{B} (\mathbb{R}) : m(A) = n (A)\}$ forms a $\lambda$-system, this should be very easy to check. For now we only have that $L\subset \mathcal{B} (\mathbb{R})$.

  3. The assumption that $m((a,b)) = n((a,b))$ imples $P\subset L$

  4. By Dynkin's $\pi - \lambda$ Theorem, we know that $$\sigma(P) \subset L,$$ since $\sigma(P) = \mathcal{B} (\mathbb{R})$, thus $L = \mathcal{B} (\mathbb{R})$, which means the two measures agree on all $B \in \mathcal{B} (\mathbb{R})$.

  • 1
    $\begingroup$ I don't find so easy proving in 2 that $L$ is closed under complements. I feel the need to use that sets $[-n,n]$ cover $\mathbb{R}$ and have finite measure. $\endgroup$
    – Anguepa
    Jul 9, 2017 at 6:59
  • $\begingroup$ In general it is not true that two measures that agree on a $\pi$-system agree on the generated $\sigma$-algebra. For example the Lebesgue measure $\mu$ and two times the Lebesgue measure $2\mu$ agree on the $\pi$-system of intervals of the form $(r,\infty)$. $\endgroup$
    – Anguepa
    Jul 9, 2017 at 7:17
  • $\begingroup$ Yes, we need finite measure, this theorem is often used in probability theory where $\mu$ takes value in $[0,1]$. $\endgroup$
    – Xiao
    Jul 10, 2017 at 22:51
  • $\begingroup$ Your proof sketch holds for two finite measures $\mu$ and $\lambda$ such that $\mu(\mathbb{R})=\lambda(\mathbb{R})$ (e.g. all probability measures). But it also holds in the case OP wants it to hold. See the answers here and here. $\endgroup$
    – Anguepa
    Jul 11, 2017 at 7:48

With your help, this is the proof that I have come up with if anyone reading this at a later time is interested.

My Solution :

First, suppose that $m(\mathbb R) = n(\mathbb R) < \infty$. We show that $m=n$ on $\mathcal B$. Consider the collection $\mathcal G = \{B \in \mathcal B: m(B) = n(B)\}$. We observe that $\mathcal G$ is a monotone class:

Suppose $A_i \uparrow A$ where $A_i \in \mathcal G$ for all $i \in \mathbb N$ and define $B_1 = A_1$, $B_2 = A_2 - A_1$, $\dots$, $B_n = A_n - \cup_{i=1}^{n-1}A_i$ so that $\cup_{i=1}^\infty A_i = \cup_{i=1}^\infty B_i$ where $\{B_i\}$ are pairwise disjoint. Notice that for $C,D \in \mathcal G$ with $C \subset D$, we have $$m(D-C) = m(D) - m(C) = n(D) - n(C) = n(D-C),$$ since $m$ and $n$ are finite measures. So $B_i \in \mathcal G$ for all $i \in \mathbb N$. Then $A \in \mathcal G$, since $$m(A) = m(\cup_{i=1}^\infty B_i) = \sum_{i=1}^\infty m(B_i) = \sum_{i=1}^\infty n(B_i) = n(\cup_{i=1}^\infty B_i) = n(A)$$

Now suppose that $A_i \downarrow A$ where $A_i \in \mathcal G$ for all $i \in \mathbb N$. Then, $$m(A) = \lim_{n\to\infty}m(A_n) = \lim_{n\to\infty}n(A_n) = n(A)$$ since $m$ and $n$ are finite measures. So $A \in \mathcal G$, which makes $\mathcal G$ a monotone class. More specifically, it makes $\mathcal G$ a monotone class that contains $\mathcal C = \{(a,b):a,b\in\mathbb R\} \cup \{\mathbb R,\emptyset\}$. Thus, $\mathcal M(\mathcal C) \subseteq \mathcal G$ where $\mathcal M(\mathcal C)$ represents the smallest monotone class containing $\mathcal C$.

Since $\mathcal C$ is a collection of subsets of $\mathbb R$ closed under finite intersections (any intersection is either an open interval or the emptyset) containing $\mathbb R$, we can apply the Monotone Class Theorem. Thus, we get that $\mathcal M(\mathcal C) = \sigma(\mathcal C) = \mathcal B$, and more importantly, $\mathcal B = \mathcal M(\mathcal C) \subseteq \mathcal G$. By definition $\mathcal G \subseteq \mathcal B$, so we have $\mathcal G = \mathcal B$ and $m = n$ on $\mathcal B$.

Now suppose that one of the measures is not finite and consider the following increasing sequence of sets $A_1 = (-1,1)$, $A_2 = (-2,2)$, $\dots$, $A_k = (-k,k)$. Define $m_k(B) = m(B\cap A_k)$ and $n_k(B) = n(B \cap A_k)$ for all $B \in \mathcal B$. Then, $m_k, n_k$ are measures for every $k \in \mathbb N$, and moreover, are finite by monotonicity of measures (they are subsets of sets with finite measure). Therefore, since $m_k(\mathbb R) = n_k(\mathbb R) < \infty$ for all, we can apply the finite case as proved above to get that $m_k = n_k$ on $\mathcal B$ for all $k \in \mathbb N$. Then for any $B \in \mathcal B$, we have $$m(B) = \lim_{k\to\infty}m(B \cap (-k,k)) = \lim_{k\to\infty}m_k(B) = \lim_{k\to\infty}n_k(B) = \lim_{k\to\infty}n(B\cap (-k,k)) = n(B).$$ Thus, $m=n$ on $\mathcal B$.

  • 1
    $\begingroup$ The Monotone Class Theorem in Wikipedia states that $\mathcal{C}$ is an algebra of sets, then the smalles monotone class containing $\mathcal{C}$ is precisely the $\sigma$-algebra generated by $\mathcal{C}$. It seems like you are using a different result. Were you? $\endgroup$
    – Anguepa
    Jul 9, 2017 at 7:22

Lemma 7.1.2. (p. 68) of Measure Theory, volume 1, Vladimir I. Bogachev:
If two finite signed Borel measures on any topological space coincide on all open sets, they coincide on all Borel sets.

Its simple proof uses:
Lemma 1.9.4. If two probability measures on a measurable space $(X,A)$ coincide on some class $E\subset A$ that is closed with respect to finite intersections, then they coincide on the $\sigma$-algebra generated by $E$.

Link to Lemma 7.1.2 A related question: If two Borel measures coincide on all open sets, are they equal?


The Borel sigma algebra on $\mathbb R$ has various equivalent generating sets: obviously the open intervals $(a, c)$ and also the half open intervals $[b, c)$ . The half open intervals are a semi-ring, and by Carathéodory a sigma finite measure (which you have here since each (finite) open interval has finite measure) defined on a semi-ring extends uniquely to a measure on the sigma-algebra generated from it.

So, if $m = n$ for each half open interval you are done.

For any half open interval $[b, c)$ there is $a \in \mathbb R $ with $a < b$.
Then $[b, c) = (a, c)\setminus (a, b)$ and $(a, b)\subset (a, c)$
So for any measure $\mu$ where $\mu( (a, b) ) $ is finite $\mu( [b, c)) = \mu((a, c)) - \mu((a, b))$
And since $m = n$ for each open interval they therefore agree on the half open intervals. $\blacksquare$


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